Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,
For 8 shot burst, average recoil force on the gun is :
So, the average recoil force on the gun during that 0.40 s burst is 45 N.
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Over the first 30.0 s, the undergoes a displacement of
(12 m/s) * (30.0 s) = 360 m
Over the next 8.00 s, the car accelerates from 12 m/s to a top speed of
12 m/s + (1.5 m/s²) * (8.00 s) = 24 m/s
and over this time interval, it is displaced an additional
(12 m/s) * (8.00 s) + 1/2 (1.5 m/s²) * (8.00 s)² = 144 m
For the last 12.0 s, the car moves at a constant speed of 24 m/s to cover a distance of
(24 m/s) * (12.0 s) = 288 m
So the car's net displacement is 360 m + 144 m + 288 m = 792 m. (The net displacement is the same as distance in this case because the car moves in only one direction.)
